试判断从键盘输入的正整数能被5和7同时整除.能则输出yes,否则输出no

importjava.io.*;importjava.util.*;classMain{publicstaticintgcd(intm,intn){while(true){if((m=m%n)==0)

intn,sum=0;cin>>n;for(inti=1;i

已修改好：#include"iostream.h"int main(){ int i; cin>>i; if(i%3=

判断是否存在《=0的值找出最大数判断两边之和大于第三边 两边之差小于第三边（a==b）==c 是错误的判断想 （a==b）返回的是1或者0 在判断==c肯定是错

dimi,j,n,tasintegern=inputbox("n=")fori=1tonifimod2=0thent=i\2forj=2toint(sqrt(t))iftmodt=othenexitf

int main(){\x09int n,max;\x09scanf("%d",&n);\x09max=n;\x09while(n>0)\x09{

#include#includemain(){floata,b,c,x,y,z;floatd;doubleS;while(1){printf("请输入三角形三边的边长数：");pri

#includeintmain(){inta;intsum=0;scanf("%d",&a);if(a

#includemain(){intn,i,j,k=0,sum=0,a[100];printf("请输入一个数字\n");scanf("%d",&n);for(i=1;i

;本程序通过编译,运行正确CodeSegmentAssumeCS:Code,DS:CodeCRequ000DHLFequ000AHKBBackequ0008H;－－－－－－－－－－－－－－－－－－－－

staticvoidMain(string[]args){inta=int.Parse(Console.ReadLine());intb=a%10;intc=(a/10)%10;intd=(a/100

inta,sum,count;doubleavg;sum=0;count=0;while(1){printf("请输入正整数（输入0结束输入）:");scanf("%d",&a);if(a==0){b

/*#include"stdio.h"main(){intx;printf("请输入一个数字:");scanf("%d",&x);if(x%3==0&&x%5==0)printf("该数字能同时被3,

#include"stdio.h"intmain(){intLength[3];for(inti=0;i

main函数：Scannerinput=newScanner(System.in);intnum=input.nextInt();if(num%3==0||num%5==0){system.out.p

publicstaticvoidmain(String[]args){Scannerscan=newScanner(System.in);while(scan.hasNextInt()){i=scan

PrivateSubCommand1_Click()a=Val(InputBox("输入a的长"))b=Val(InputBox("输入b的长"))c=Val(InputBox("输入c的长"))If

//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"boolprime(intn){inti;if(n>2&&!(n&1)||n

intn;scanf("%d",&n);if(n%5==0&&n%7==0){printf("YES");}else{printf("NO");}

DIMASINTEGERA(5),C(9)FORI=1TO5INPUTA(I)NEXTIFORI=1TO5T=A(I)DOWHILET0T=T\10S=S+1LOOPT=A(I)DIMB(S)ASIN